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A hydraulic lift is shown in the figure. The radii of the movable pistons $P_1$ and $P_2$ are $2 \mathrm{~m}$ and $5 \mathrm{~m}$ respectively. If a block of mass $x$ is placed on $P_2$ then the minimum mass that should be kept on $P_1$ to lift the block on $P_2$ is
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Verified Answer
The correct answer is:
$0.16 x$
Given $r_1=2 \mathrm{~m}, r_2=5 \mathrm{~m}$
According to Pascal's law
$$
\begin{aligned}
p_1 & =p_2 \\
\frac{F_1}{A_1} & =\frac{F_2}{A_2} \\
\Rightarrow \quad \frac{m_1 g}{\pi r_1^2} & =\frac{m_2 g}{\pi r_2^2}
\end{aligned}
$$
$\begin{array}{rlrl}\Rightarrow & & \frac{m_1}{r_1^2} & =\frac{m_2}{r_2^2} \\ \Rightarrow & \frac{m_1}{2^2} & =\frac{x}{5^2} \\ \Rightarrow & & m_1 & =\frac{4}{25} x \\ & & =0.16 x\end{array}$
According to Pascal's law
$$
\begin{aligned}
p_1 & =p_2 \\
\frac{F_1}{A_1} & =\frac{F_2}{A_2} \\
\Rightarrow \quad \frac{m_1 g}{\pi r_1^2} & =\frac{m_2 g}{\pi r_2^2}
\end{aligned}
$$
$\begin{array}{rlrl}\Rightarrow & & \frac{m_1}{r_1^2} & =\frac{m_2}{r_2^2} \\ \Rightarrow & \frac{m_1}{2^2} & =\frac{x}{5^2} \\ \Rightarrow & & m_1 & =\frac{4}{25} x \\ & & =0.16 x\end{array}$
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